There are three source of sound of equal intensities with frequencies 400, 401 and 402 Hz. The no. of beats per second is :

Options

A. 0
B. 1
C. 2
D. 3

Answer

B

Explanation

We known that y = r sin `2 pi v t`.
` therefore y_(1) = r sin 2pi 400 t , y_(2) = r sin 2 pi 401 t `
`y_(3) = r sin 2pi 402 t.`
Now y = `y_(1) + y_(3) + y_(2)`
y = r sin `2pi` 400t + r sin `2pi` 402 t + r sin `2pi` 401 t.
y = r sin `2 pi ((402 - 400)/(2)) t cos 2pi ((402 - 400)/(2) ) + r sin 2 pi 401 t`.
`y = 2r sin 2pi 401 t cos 2pi t + r sin 2 pi 401 t`.
y = `[2 cos 2 pit + 1] r sin 2 pi 401 t`.
Now Amplitude of resultant wave is
` A = (2 cos 2pi t + 1) r. `
For A to be maximum , cos `2 pi t = + 1 `
`rArr " " 2 pi t = 2 n pi, n = 0,1,2, t = n `
Amplitude will be maximum at time t = 0 , 1,2,3 ... sec.
Time interval between successive maximum = 1 sec.
`therefore` beat frequency = 1 Hz.
hence the correct choice is (b) .